In this Section I show that the result of our commitment game displayed in Figure
2.2.1 on page 66 is the non-commitment outcome, that is population A players will
choose to bluff instead of commit and this is met with the Hard response by population
B players. Furthermore this result is reasonably robust to a change in the model,
allowing B to discover bluffs with small probability.
Perfect disguise
To start with I assume perfect disguise. That is A is perfectly adept at bluffing
commitment so that B receives no information whatsoever about the action chosen
by A. This has already been discussed in Section 2 and fully represented in Figure
2.2.1 on page 66.
instead of committing and soB weakly dominatesC. In addition to the obvious Nash Equilibrium (B, H), there is also a connected component of Nash Equilibria giving
the commitment outcome (c,1−c): B plays S and A plays C with probability at
least 1−c−ww. The absorbing states of the population game correspond to these Nash Equilibria: xA, xB= ((x,1−x),(1,0)) : x∈ c−w 1−w,1 , NAx∈Z ∪ {((0,1),(0,1))}
If the process is in any one of these states, it will require a mutation to leave that
state. The connected component with S being played are linked by a chain of single
population A mutations. From any member of this component, we require just one
population B mutation to get to the (B, H) equilibrium of ((0,1),(0,1)). However,
as long as population A is large enough, it requires more than one mutation to get
from ((0,1),(0,1)) to a member of the connected component. This gives the following result:
Theorem 26. For any population sizesNA, NBsatisfyingNA ≥ 1−w
c−w and N
B ≥1,
the LRE is x(B,H) = ((0,1),(0,1))
Proof. The result follows from Ellison’s radius-coradius theorem. CRx(B,H) = 1 since a single Hard agent in populationBis enough to induce all populationA agents to bluff, which in turn induces all populationBto to act hard. WhileRx(B,H)>1,
since NA ≥ 1−w
c−w ensures that one population A mutation is not enough to entice
agents in population B to switch to soft.
An important note, in light of the Kim and Wong [25] critique, is that these are the only strategies considered. Consider the following Example:
Example 27. Suppose that the method A agents use to claim commitment is em- ploying a bargaining agent allegedly under contract not to accept less than c, where
S H bribe
C (c,1−c) (0,0) (2 +w,−1−w)
B (c,1−c) (w,1−w) (w,−1−w)
Table 2.3.1: Adding “bribe” strategy
the terms of the contract compel the bargaining agent to pay 2 (in payoff units) to A
should he break this. Now, one option available toB, albeit a deliberately nonsensical
one, is to bribe the bargaining agent 2 units to accept w. Then denoting this new
strategy by “bribe”, the new payoff table is given in Table 27 on page 83
Note that this option to bribe the bargaining agent with 2 units will always give
B a payoff of −1−w < 0 and so is strictly dominated and hence should never be
played. But the fact that agents in populationB could mutate to this strategy affects the results by reducing Rx(B,H). If NB ≤ 2+w
w then from x
(B,H) a single mutation of a populationBagent from “H” to “bribe” is sufficient to makeC the best response for population A agents and thus allow us to reach state x(C,S) = ((1,0),(1,0)). So now we can connect the set of absorbing states via a cycle with resistance 1 between any two adjacent states. Hence7 they are all stochastically stable.
I take the view that in this scenario the correct solution is obtained by not allowing the strategy “bribe” since it clearly is not a sensible strategy forBto consider thinking about players mutating through making mistakes, it is really easy to comprehend how an agent might mutate between “Soft” and “Hard” since both strategies are in some contexts sensible. Indeed, such a mutation could be from a mis-reading of the current population state. However it is far harder to make an argument for why an agent would mutate to “bribe”, since this is guauranteed to worse than either of the other two strategies. One way to model this is to allow for state dependent mutation rates as in [5] and make such mutations far less likely, but then that raises the question how unlikely to make them. Then there is also the same issue with any other dominated
7This is fairly obvious. Either directly apply the Freidlin and Wentzell tree surgery or see for
strategies that players have. So, with this in mind, I think it is more appropriate to proceed by restricting the strategy space to “sensible” strategies.
This consideration of restricting the strategy sapce also came into consideration when forming this game displayed in Figure 1. As noted in Section 2 the game represented in Figure 1 can be thought of as a truncation since it was assumed au- tomatically that A would respond optimally to the offer of B. There is a literature applying stochastic stability to extensive form games, see [10], [18], [21], [26], [35] and generally they show that the presence of such suboptimal replies can make a difference if agents are allowed to mutate toward such suboptimal replies. Once again I argue for restricting the strategy space based on what strategies are sensible. It seems hard
to justifyA wanting to turn down an offer from B in payoff terms, but also when we
consider the context of what the strategies “Commit” and “Bluff” mean. The whole point of bluffing instead of committing is to then be able to acceptw if B acts Hard.
So in light of this, it would seem very odd behaviour to then turnw down.
Imperfect disguise
Now I drop the asssumption that a bluff is never discovered, so that now there is some incentive forAto commit rather than bluff against an opponent choosing a soft strategy. If we suppose a bluff is discovered with probabilityλ∈(0,1) then the game is represented in Figure 2.3.1 on page 85.
Note that once again this could be viewed as a slightly truncated version of the game. On top of the truncation described in Figure 2.2.1 on page 66, I have also
truncated the game slightly by assuming that after A bluffs and nature reveals the
bluff, B automatically chooses Hard and A accepts w, that is players follow the
backwards induction equilibrium in this subgame. I argue that it is fairly obvious given the interpretation of the game that this is the only sensible play for both players. The reason for truncating the game in this way is the same as before.
A C B Nature B S H S H (0,0) (𝑐, 1 − 𝑐) (𝑐, 1 − 𝑐) (𝑤, 1 − 𝑤) λ 1 −λ (𝑤, 1 − 𝑤) S H C B (𝑐, 1 − 𝑐) (0,0) (𝑤, 1 − 𝑤) (𝑟, 1 − 𝑟) B A 𝑟 =λ𝑤 + 1 −λ𝑐 In normal form In extensive form After observing a bluff, B holds firm, forcing A to back down
Figure 2.3.1: Game under imperfect disguise assumption
There is a second interpretation of this game, which has the same game tree
structure as Figure 2.2.1 on page 66. Suppose that post-agreement, Bmight discover
A was bluffing and be able to use this to renegotiate. For example, returning to the firm-union example, suppose the union bluffs and the firm concedes to the bluff, but later discovers this was a bluff. Although the union gets its way with the current pay deal, the firm may be able to use the knowledge of the bluff to its advantage during future negotiations.
Notice that in this new payoff matrix given in Figure 2.3.1 on page 85,B no longer
weakly dominates C and so both (C, S) and (B, H) are now strict Nash equilibria.
This means that x(C,S) = ((1,0),(1,0)) and x(B,H) = ((0,1),(0,1)) are the only absorbing states8, and for large populations, both states will require several mutations to escape from and so we must apply the mutation counting methods discussed in Section 2. We findα= c−wr+w andβ = 1−rc+−wr−w are the cutoffs determining the basins
8It is posible to have a mixed strategy Nash Equilibrium be an absorbing state if the equilibrium
mixtures are compatible with the integer problem from the population sizes. But generically this will not be the case, and even when such an absorbing state does exist it will be very unstable to mutations and not a candidate for being stochastically stable.
of attraction and so applying Theorem 23, we obtain
Theorem 28. If r >max{c−w,1−(c−w)} (ie λ < minnc−ww,2cc−−ww−1o) and both populations have either an even or sufficiently large number of agents, then x(B,H) is
the LRE.
If r < min{c−w,1−(c−w)} (ie λ >maxnc−ww,2c−c−ww−1o) and both populations have either an even or sufficiently large number of agents, then x(C,S) is the LRE. Proof. Note that α > 1/2 ⇐⇒ r > c−w and β > 1/2 ⇐⇒ r > 1−(c−w) and the result follows from 25
If the population sizes are equal and arbitrarily large so that stochastic stability coincides with risk dominance.
Theorem 29. Let N1 = N2 = N → ∞. If r > c−c2+w2
1−c+w (ie λ < w
1−c+w) then x
(B,H)
is the LRE.If r < c−1−c2c++ww2 (ie λ > 1−wc+w) then x(C,S) is the LRE.
Proof. Some elementary algebra shows that min{α, β} > min{1−α,1−β} ⇐⇒
r > c−1−c2c++ww2. Then the result follows from 25
This means that, as long as A is sufficiently adept at bluffing so that the prob- ability λ of the bluff being discovered is reasonably small, the outcome (B, H) will still prevail. On the other hand, if the probability of a bluff is likely to be discovered as such then the commitment outcome of (C, S) prevails. This makes good intuitive
sense, as when λ= 1 we are in the perfect information case where B knows whether
A is bluffing or committed. In this case it is well established that A can commit.
Note that while, the ability to bluff well may at first sight seem a strength, it is in
fact a weakness since B then expect A to use this ability and thus chooses the Hard
strategy.
B
S H
A C (0.8,0.2) (0,0)
B (0.75,0.25) (0.5,0.5)
Table 2.3.2: Imperfect disguise example with x(C,S) being the LRE Example 30. Let payoffs be given by Table 2.3.2 on page 87.
The only candidates for stochastically stable sets are the absorbing states of the process without mutations, these x(C,S) and x(B,H). The process will naturally move away from any other state very quickly, since doing so does not require any mutations. So we are interested in the relative time spent in the x(C,S) state compared to that in thex(B,H) state. This is determined by the number of transitions needed to transition between the two.
To get from the x(C,S) state to the x(B,H) state requires either 4/9 of population A
mutating from C toB or 1/11of population B mutating from S toH.
To get from the x(B,H)) state to the x(C,S) state requires either 5/9 of population
A mutating from B to C or 10/11 of population B mutating from H to S.
So as ε → 0, it becomes infinitely more likely to transition from x(C,S) to x(B,H) state than vice-versa, and thus x(B,H) is the uniquely stochastically stable state.